Lecture 5
Continuity and topology
Many results on continuous functions from a metric space to another metric space are similar to those from a subset of the reals to another subset of the reals. We simply use the metric in places of the distance normally defined by the absolute value.
Here is an example. Let from the metric space
to the metric space
be continuous at a point
. Then for any
there is
as in the definition of continuity. Since
we can choose some
such that
for all
, and this means that
for all
. Hence
We can also prove the converse statement. To this end we assume that but
is not continuous at
. The latter means that we can find some
and some
such tha
yet
. Clearly this is a contradiction.
So we have proved a useful theorem concerning continuity.
Theorem. A function from a metric space
to a metric space
is continuous at a point
if and only if
for every sequence
which converges to
.
In fact there is an equivalent definition of a continuous function of a metric space (often used in topology) due to the following theorem
Theorem. A mapping of a metric space
into a metric space
is continuous on
if and only if
(the inverse image) is open in
for every open set
in
There is a discussion on how to write the proof and here. What do you think?
Since a set is closed if and only if its complement is open and for every
we have
Corollary. A mapping of a metric space
into a metric space
is continuous on
if and only if
(the inverse image) is closed in
for every closed set
in
Caution: The continuity of is not equivalent to the requirement that the image
of an open set
be open in
. An example is
,
Here
, which is not open in
.
Example.
![f(x)=x^2. f(x)=x^2.](https://kurser.math.su.se/filter/tex/pix.php/53b8895d85195de9ecabe106cf45b925.png)
![f f](https://kurser.math.su.se/filter/tex/pix.php/8fa14cdd754f91cc6554c9e71929cce7.png)
- that the inverse image of an open interval is open.
- that the inverse image of a closed interval is closed.
![](http://www.mathcs.org/analysis/reals/cont/graphics/topcont1.gif)
f(x) = x2
First, let's look at the inverse images of an open interval:
- If
then the inverse image of the open interval
is
. In particular, the inverse image is open.
- If
, then the inverse image of the open interval
is
, which is again open.
- If
, then the inverse image of the open interval
is again open (which set is it ?)
But it is now obvious that the inverse image of closed intervals is again a closed set (note that the empty set is both open and closed).
Hence, we have proved that the function is continuous, avoiding the tedious epsilon-delta proof. ♦
Now we use the theorem to prove the composition function of two continuous functions is continuous. More precisely,
Theorem. Suppose that are metric spaces,
,
maps
into
,
maps the range of
,
into
, and
is the mapping of
into
defined by
(
). If
are continuous on
and
respectively, then
is continuous.
Proof. Let be open. The key observation here is
. (This can be proven by definition.) But
is open for some open set
in
. Then
is open in
, since
and
are continuous. Therefore
is continuous. ◊